\chapter{Maxwell Equations}
选择二维情况，有如下的偏振矢量
\begin{subequations}
\begin{eqnarray}
{\bf E}&=&E_z(r,\theta)\hat{\bf z} \\
{\bf H}&=&H_r(r,\theta))\hat{ \bf r}+H_\theta(r,\theta)\hat{\bm
\theta}
\end{eqnarray}
\end{subequations}
对上式求旋度
\begin{subequations}
\begin{eqnarray}
\nabla\times{\bf E}&=&\frac{1}{r}\frac{\partial E_z}{\partial\theta}{\hat{\bf r}}-\frac{\partial E_z}{\partial r}{\hat{\bm\theta}}\\
\nabla\times{\bf H}&=&\left[\frac{1}{r}\frac{\partial}{\partial
r}(rH_\theta)-\frac{1}{r}\frac{\partial
H_r}{\partial\theta}\right]{\hat{\bf z}}
\end{eqnarray}
\end{subequations}
因为在柱坐标系下，$\overline{\overline\mu}$是对角的，所以Maxwell方程组中电场$\bf
E$的旋度
\begin{subequations}
\begin{eqnarray}
&&\nabla\times{\bf E}=i\omega{\bf B} \\
&&\frac{1}{r}\frac{\partial E_z}{\partial\theta}{\hat{\bf
r}}-\frac{\partial E_z}{\partial
r}{\hat{\bm\theta}}=i\omega\mu_rH_r{\hat{\bf r}}+i\omega\mu_\theta
H_\theta{\hat{\bm\theta}}
\end{eqnarray}
\end{subequations}
所以$\bf H$的各个分量可以写为：
\begin{subequations}
\begin{eqnarray}
H_r=\frac{1}{i\omega\mu_r}\frac{1}{r}\frac{\partial
E_z}{\partial\theta } \\
H_\theta=-\frac{1}{i\omega\mu_\theta}\frac{\partial E_z}{\partial r}
\end{eqnarray}
\end{subequations}
同样地，在柱坐标系下，$\overline{\overline\epsilon}$是对角的，所以Maxwell方程组中磁场$\bf
H$的旋度
\begin{subequations}
\begin{eqnarray}
&&\nabla\times{\bf H}=-i\omega{\bf D}\\
&&\left[\frac{1}{r}\frac{\partial}{\partial
r}(rH_\theta)-\frac{1}{r}\frac{\partial
H_r}{\partial\theta}\right]{\hat{\bf
z}}=-i\omega{\overline{\overline\epsilon}}{\bf
E}=-i\omega\epsilon_zE_z{\hat{\bf z}} \\
&&\frac{1}{r}\frac{\partial}{\partial
r}(rH_\theta)-\frac{1}{r}\frac{\partial
H_r}{\partial\theta}=-i\omega\epsilon_zE_z
\end{eqnarray}
\end{subequations}
由此我们可以得到关于$E_z$的波函数方程：
\begin{eqnarray}
\frac{1}{\mu_\theta\epsilon_z}\frac{1}{r}\frac{\partial}{\partial r}
\left(r\frac{\partial E_z}{\partial r}\right)+
\frac{1}{\mu_r\epsilon_z}\frac{1}{r^2}\frac{\partial^2E_z}{\partial\theta^2}
+\omega^2 E_z=0
\end{eqnarray}
